Question

The mean time taken by all participants to run a road race was found to be 220 minutes with a standard deviation of 30 minutes. Using Chebyshev's theorem, find the minimum percentage of runners who ran this road race in

a. 130 to 310 minutes: %

b. 70 to 370 minutes: %

c. 100 to 340 minutes: %

Answer 1

ANSWER:

Given that,

if X be a random variable with mean u and standard deviation s then by Chebyshev's theorem

P[u-ks<X<u+ks]>=1-1/k²

here u=220 minutes s=30 minutes and X: time taken by participants to run a road race

a) u-ks=130 u+ks=310 so k=(310-220)/30=3

so P[130<X<310]=P[220-3*30<X<220+3*30]>=1-1/3²=0.88888

hence the minimum percentage is 88.88%

b)

u-ks=70 u+ks=370 so k=(370-220)/30=5

hence P[70<X<370]=P[220-5*30<X<220+5*30]>=1-1/5²=0.96

so minimum percentage is 96%

c)

u-ks=100 u+ks=340 so k=(340-220)/0=4

hence P[100<X<340]=P[220-4*30<X<220+4*30]>=1-1/4²=0.9375 or 0.94

so minimum percentage is 94%